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Q5(iii):
Evaluate the determinant: (iii) $\begin{vmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{vmatrix}$
Solution :
Step 1: Establishing the Determinant Structure
To evaluate the determinant of a $3 \times 3$ matrix $A$, we apply the Laplace expansion method. Let $A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}$. The determinant $|A|$ is calculated by expanding along any row or column. We shall expand along the first row ($R_1$).
The formula for expansion along $R_1$ is: $|A| = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})$
Step 2: Identifying Matrix Elements
Given the determinant $\Delta = \begin{vmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{vmatrix}$, we identify the coefficients:
- $a_{11} = 0, a_{12} = 1, a_{13} = 2$
- $a_{21} = -1, a_{22} = 0, a_{23} = -3$
- $a_{31} = -2, a_{32} = 3, a_{33} = 0$
Step 3: Calculating Minors and Expansion
Substituting the values into the expansion formula [Per the Definition of Determinant of a Matrix]:
$\Delta = 0 \begin{vmatrix} 0 & -3 \\ 3 & 0 \end{vmatrix} - 1 \begin{vmatrix} -1 & -3 \\ -2 & 0 \end{vmatrix} + 2 \begin{vmatrix} -1 & 0 \\ -2 & 3 \end{vmatrix}$
Calculating the $2 \times 2$ determinants:
- First term: $0 \cdot [(0)(0) - (-3)(3)] = 0 \cdot [0 - (-9)] = 0$
- Second term: $-1 \cdot [(-1)(0) - (-3)(-2)] = -1 \cdot [0 - (6)] = -1 \cdot (-6) = 6$
- Third term: $2 \cdot [(-1)(3) - (0)(-2)] = 2 \cdot [-3 - 0] = 2 \cdot (-3) = -6$
Step 4: Algebraic Summation
$\Delta = 0 + 6 + (-6)$
$\Delta = 6 - 6$
$\Delta = 0$
Final Solution: The value of the determinant is $0$.
More Questions from Class 12 Mathematics Determinants EXERCISE 4.1
- Q1: Evaluate the determinant: $\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}$
- Q2(i): Evaluate the determinant: (i) $\begin{vmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix}$
- Q2(ii): Evaluate the determinant: (ii) $\begin{vmatrix} x^2 - x + 1 & x - 1 \\ x + 1 & x + 1 \end{vmatrix}$
- Q3: If A = $\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$, then show that $| 2A | = 4 | A |$
- Q4: If A = $\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix}$, then show that $| 3 A | = 27 | A |$
- Q5(i): Evaluate the determinant: (i) $\begin{vmatrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{vmatrix}$
- Q5(ii): Evaluate the determinant: (ii) $\begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix}$
- Q5(iv): Evaluate the determinant: (iv) $\begin{vmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{vmatrix}$
- Q6: If A = $\begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix}$, find $| A |$
- Q7(i): Find values of $x$, if (i) $\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}$
- Q7(ii): Find values of $x$, if (ii) $\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}$
- Q8: If $\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}$, then $x$ is equal to
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